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A Rolling ball and generalized co-ordinates

  1. Mar 29, 2017 #1
    Consider a sphere constrained to roll on a rough FLAT HORIZONTAL surface. A book on classical mechanics says it requires 5 generalized co-ordinates to specify sphere's configuration: 2 for its centre of mass and 3 for its orientation.

    I did not understand why 3 for orientation. I guess only 2 are needed: one altitude and one azimuth i.e. ## \theta## and ## \phi ##. Since radius is fixed, ## r ## is not needed.

    So only 2+2 = 4 generalized co-ordinates are needed as per my understanding.

    I assume book is right. So what is wrong in my understanding ?
     
    Last edited: Mar 29, 2017
  2. jcsd
  3. Mar 29, 2017 #2

    PeroK

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    In general, for a three dimensional rigid body you need three angles to define its orientation. For example, a rotation about three axes.

    Alternatively you can define a unit vector in any direction (your two polar and azimuthal angles) and a third angle that defines the rotation about that vector.
     
  4. Mar 29, 2017 #3
    But consider a fixed point on the sphere. We need 4 co-ordinates to uniquely decide its location on the sphere.
    (x,y) to locate COM of the sphere and altitude & azimuth to locate the point on the surface of the sphere.
    (2 co-ordinate systems are used : XY plane system and spherical co-ordinate system. )
    Since sphere is rigid body, if we rotate sphere about the axis passing through our fixed point and the COM,
    orientation of the sphere will change (since our fixed point is connected to other points inside the sphere). Isn't it ?
     
  5. Mar 29, 2017 #4

    PeroK

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    That gives 5 coordinates by my counting: 2 for the COM; 2 for the reference point on the surface (imagine a dot on the sphere); and 1 for a rotation about the line joining the COM and the dot.
     
  6. Mar 29, 2017 #5
    But when you rotate sphere about line joining COM and the dot, co-ordinates of other points (on & inside) of the sphere are changing.
    If I move the sphere from (x1,y1) to (x2,y2) (while changing at the same time polar and azimuth angles), motion is complete.
    Dot on the sphere is not in motion now. Why do we need further rotation along the line joining the COM and the dot.
     
  7. Mar 29, 2017 #6

    PeroK

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    Do you follow the point about the orientation of the sphere in general?

    Is your question about the specific situation where the ball must roll without slipping?

    If so, the ball can get from one point to another by a path of any length - it doesn't have to roll in a straight line. This means that, say, the ball could return to its starting point with any orientation - not just the one it started with.
     
  8. Mar 29, 2017 #7
    Yes, ball must roll without slipping. So do you mean 3 angles (in general) will change when ball is in motion?
    These 3 angles being axes of rotation, one rotation about X axis, one rotation about Y axis & one rotation about Z axis.
    (This XYZ system being superimposed on spherical co-ordinate system and both having same origin which is COM.)
    Is this correct ?
     
  9. Mar 29, 2017 #8

    PeroK

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    Please answer my first question:

    Do you follow the point about the orientation of the sphere in general?

    Yes or no.
     
  10. Mar 29, 2017 #9
  11. Mar 29, 2017 #10

    PeroK

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    If we forget about the movement of the COM, do you think that a 3D rigid body has two or three degrees of freedom in respect of rotations?
     
  12. Mar 29, 2017 #11
    3 degrees of freedom i.e. 3 axes of rotations. And 3 axes being orthogonal to each other.
    Each axis passing through COM (their point of intersection).
     
  13. Mar 29, 2017 #12

    PeroK

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    So, a sphere has three degrees of freedom with respect to rotations, hence six degrees of freedom with respect to position and orientation, which reduces to five if you take away the freedom to move off a plane.
     
  14. Mar 29, 2017 #13
    Fantastic! Now I understood. Thanks for clear & logical answer.

    Now my next doubt is : If we consider motion of the sphere restricted to flat,horizontal plane (without slipping), can all these 5 generalized co-ordinates vary independently of each other ?
     
  15. Mar 29, 2017 #14

    PeroK

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    That's a good question. The answer is yes, but I must admit I don't know how to prove it. Try looking on line. Just "rolling ball coordinates" should do it.

    Imagine two reference points on the sphere. As long as they are not antipodal, they define the orientation. Perhaps a black dot initially at the north pole and red dot on the equator.

    The hand-wavy argument is that there are infinitely many paths that will get the sphere back to where it started with the black dot back at the north pole - and the red dot is not necessarily where it started.

    A formal proof would have to show that all points on the equator are possible.
     
  16. Mar 29, 2017 #15
    But my book says these 5 generalized co-ordinates can NOT all vary independently. When the sphere rolls on perfectly rough flat,horizontal surface, at least 2 co-ordinates (out of 5) must change. Hence degrees of freedom for this case is < 5; but number of generalized co-ordinates is 5. Hence it is a non-holonomic system. Do you agree with this ?
     
  17. Mar 29, 2017 #16

    PeroK

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    Yes, because you can describe the motion of the sphere by two parameters at any time: it's simply rolling in a direction in the x-y plane. A point mass would have two degrees of freedom and two coordinates.

    The configuration of the sphere depends, however, on the path it took. That's relatively easy to see.

    To prove that any configuration is possible and hence the full five coordinates are needed is not so easy to see.
     
  18. Mar 30, 2017 #17
    Free rigid body (rigid body that is flying in space) has 6 degrees of freedom: for example three Euler angles and the Cartesian coordinates of its center. Now put the ball on a rough surface; and let ##S## be the center of the ball and ##A## be a point of the ball that is in contact with the surface at this moment. Then we have
    $$\boldsymbol v_A=\boldsymbol v_S+\boldsymbol\omega\times\boldsymbol{SA}=0.$$ These are three scalar equations of constraint. So the number of degrees of freedon is ##6-3=3##. But these are nonholonomic constraints: we can not describe the position of the ball by three generalized coordinates.

    UPD
     
    Last edited: Mar 30, 2017
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